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lisov135 [29]
3 years ago
13

A bumper car with a mass of 240 kg is moving to the right with a velocity of 2.3

Physics
1 answer:
Sauron [17]3 years ago
7 0

Answer:

See the explanation below

Explanation:

To solve this problem we will take the definition of linear momentum, which tells us that momentum is equal to the product of mass by the velocity vector. Since velocity is a vector, we will take the right-hand movement as positive and the left-hand movement as negative, the left-hand members are taken as before the collision and the right-hand members as after the collision

ΣM1 = ΣM2

(m1*v1) + (m2*v1) = (m1*v2) + (m2*v2)

V1 = velocity before the collision [m/s]

V2 = velocity after the collision [m/s]

m1 and m2 = mass of the vehicles [kg]

Therefore:

(240*2.3) - (260*2.7) = - (240*2.9) + (260*2.1)

Resolving this arithmetic operation we will have:

720 - 702 = - 696 + 546

- 150 = - 150

We can see that before the crash and after the crash the momentum is preserved

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Coulomb's law is expressed mathematically as
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Force between two charges  = 

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4 0
3 years ago
Nerve cells transmit electric signals through their long tubular axons. These signals propagate due to a sudden rush of Na+ ions
iren [92.7K]

Answer:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

Explanation:

For this case we have the following info given:

Number of Na+ ions 5.7 x10^{11} ions

Each ion have a charge of +e and the crage of the electron is 1.6 x10^{-19}C

The time is given t = 7 ms if we convert this into seconds we got:

t = 7ms * \frac{1s}{1000 ms}= 0.007s

Now we can use the following formula given from the current passing thourhg a meter of nerve axon given by:

Q = Ne

Where N represent the number of ions, e the charge of the electron and Q the total charge

If we replace on this case we have this:

Q= 5.7x10^{11} * (1.6 x10^{-19}C) = 9.12x10^{-8} C

And from the general definition of current we know that:

I =\frac{Q}{t}

And since we know the total charge Q and the time we can replace:

I = \frac{9.12x10^{-8} C}{0.007 s}= 1.30285 x10^{-5} \frac{C}{s}=1.30285 x10^{-5} A = 13.02 \mu A

The current during the inflow charge in the meter axon for this case is 13.02 \mu A

3 0
3 years ago
Two cars are traveling along a straight road. Car A maintains a constant speed of 95 km/h and car B maintains a constant speed o
natulia [17]

Answer

given,

Speed of car A = 95 Km/h

                         = 95 x 0.278 = 26.41 m/s

Speed of Car B = 121 Km/h

                         = 121 x 0.278 = 33.64 m/s

Distance between Car A and B at t=0 = 41 Km

a) Distance travel by car B

   d = 26.41 t + 41000

speed of the car A = 33.64 m/s

distance = s x t

26.41 t + 41000 = 33.64 x t

7.23 t = 41000

t = 5670.82 s

time taken by Car B to cross Car A is equal to t = 5670.82 s

distance traveled by car A

D = s x t  = 26.41 x 5670.82 = 149766.25 m = 149.76 Km

b) distance travel by the car B in 30 s after overtaking car A

   D' = s x t = 33.64 x 30 = 1009.2 m = 1 Km  

8 0
3 years ago
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